Problem: A local gift shop sold bags of candy and cookies for Halloween. Bags of candy cost $$8.00$, and bags of cookies cost $$3.50$, and sales equaled $$45.00$ in total. There were $3$ more bags of cookies than candy sold. Find the number of bags of candy and cookies sold by the gift shop.
Let $x$ equal the number of bags of candy and $y$ equal the number of bags of cookies. The system of equations is then: ${8x+3.5y = 45}$ ${y = x+3}$ Since we already have solved for $y$ in terms of $x$ , we can use substitution to solve for $x$ and $y$ Substitute ${x+3}$ for $y$ in the first equation. ${8x + 3.5}{(x+3)}{= 45}$ Simplify and solve for $x$ $ 8x+3.5x + 10.5 = 45 $ $ 11.5x+10.5 = 45 $ $ 11.5x = 34.5 $ $ x = \dfrac{34.5}{11.5} $ ${x = 3}$ Now that you know ${x = 3}$ , plug it back into $ {y = x+3}$ to find $y$ ${y = }{(3)}{ + 3}$ ${y = 6}$ You can also plug ${x = 3}$ into $ {8x+3.5y = 45}$ and get the same answer for $y$ ${8}{(3)}{ + 3.5y = 45}$ ${y = 6}$ $3$ bags of candy and $6$ bags of cookies were sold.